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Pell Mell

Submitted by Billy from Montreal, QC, 3/4/2000. Original answer by Carl Johan Ragnarsson; this article by Allen Stenger.

Show that for any n>=1 the number (sqrt(2)-1)^n can be written as the difference of the square roots of two consecutive integers.

Hint 1

Work out the first few powers and verify the result for them. Do you see any patterns?

Hint 2

Here are the first 5 powers:

first 5 powers of sqrt(2) - 1

If we turn this around we get a nicer pattern:

first 5 powers of 1 - sqrt(2)

Based on these examples, it appears that the result is correct, and in fact the first square root is the square root of a perfect square. Do you see any relationships between the numbers for different powers?

Hint 3

Let's write

A_n + B_n sqrt(2)

From the table it appears that each A value is about twice the previous value, and in fact it appears that the difference is

recursion for A_n

Also it appears that

recursions for A_n and B_n

This table swarms with recursions! See if you can prove some of them.

Hint 4

Since the thing we are interested in (a power) has a recursive definition, we can use that to prove the recursions:

proof of recursions

and therefore

recursions for A_n and B_n

Now see if you can prove the difference in the squares is 1 or -1.

The Rest of the Solution

Again we proceed by recursion: If it is true for n, then

proof it is the difference of consecutive squares

Is This a Surprising Result?

What special property does sqrt(2)-1 have, that its powers are always the difference of the squares roots of two consecutive integers? It's easy to see that any power of sqrt(2)-1 has the same property; are there other kinds of numbers with this property? Yes! Here are some even more unlikely numbers with this property:

more magical numbers

What's going on here? We can get a better understanding by approaching the problem from the other end. In our proof we dealt with an equation of the form

A^2 - N B^2 = 1

(for N=2). We can factor this as

(A - B sqrt(N))(A + B sqrt(N) = 1

and raise each term to a power

(A - B sqrt(n))^n(A + B sqrt(N)^n = 1

When you multiply these powers out, you will see that the terms have the form

(C - D sqrt(N))(C + D sqrt(N)) = 1

that is,

C^2 -N D^2 = 1

In other words, any solution of this equation leads to additional solutions, by raising a certain expression to a power, and all powers of this expression are the difference of the square roots of two consecutive integers.

The equation

A^2 - N B^2 = 1

is called Pell's equation, and there is a great deal of Pell lore in Number Theory. We've seen just a little bit of this lore here; there is much more in connection with recursions, rational approximations to square roots, and continued fractions. See Shanks's book for more.

References

  • Edmund Landau, Elementary Number Theory, AMS Chelsea, 1966, pp. 76-84. Many books on number theory cover Pell's equation; this book gives an especially clear and simple development.
  • Daniel Shanks, Solved and Unsolved Problems in Number Theory, AMS Chelsea, 1993. Pell's equation is spread all through Chapter III, but see especially Theorem 77 and the surrounding pages.
  • Click here to view the original problem submitted by Billy.

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